I Best Derivation of E=mc^2 - Fdx = V^2 dm

[JLT]

TL;DR Summary

best derivation

F= V dm/dt = V (dm/dx)(dx/dt) = V^2 (dm/dx)
Fdx = V^2 dm
E = m v^2. ,mass flowing with constant velocity

If velocity is changing rather than mass, then E = 1/2 m V^2

Ok, joking aside, what is the best derivation of E= mc^2 ?

 

[malawi_glenn]

What does "best" mean?

(one dimensional case) start with relativistic momentum $p = \gamma(u) m u$
Force: $F_\text{res} = \dfrac{\mathrm{d} p}{\mathrm{d} t} = \dfrac{\mathrm{d} }{\mathrm{d} t} (m\gamma(u) u) = ma\gamma(u)^3$ where $a$ is the acceleration.
$\displaystyle \int a \gamma(u)^3 \mathrm{d} x = c^2\gamma(u) + C$
Work: $\displaystyle W_\text{tot} = \int_{x_1}^{x_2} F_\text{res} \mathrm{d} x = m\int_{x_1}^{x_2} a \gamma^3 \mathrm{d} x = mc^2\cdot \gamma (u(x_2)) -mc^2\cdot \gamma(u(x_1)) =mc^2\cdot \gamma (u_2) -mc^2\cdot \gamma(u_1)$
Work is defined as change in kinetic energy, and kinetic energy $E_\text{k}$ at a given instant is defined as the work needed to bring particle to a certain speed $u_2$ from rest $u_1=0$.
Thus, we have $E_\text{k} = m\gamma(u) c^2 - mc^2$.
The last term $mc^2$ can be interpreted as "rest energy" and therefore the $m\gamma(u) c^2$ as the total energy $E_\text{tot}$

 

[JLT]

What does "best" mean?

(one dimensional case) start with relativistic momentum $p = \gamma(u) m u$
Force: $F_\text{res} = \dfrac{\mathrm{d} p}{\mathrm{d} t} = \dfrac{\mathrm{d} }{\mathrm{d} t} (m\gamma(u) u) = ma\gamma(u)^3$
$\displaystyle \int a \gamma(u)^3 \mathrm{d} x = c^2\gamma(u) + C$
Work: $\displaystyle W_\text{tot} = \int_{x_1}^{x_2} F_\text{res} \mathrm{d} x = m\int_{x_1}^{x_2} a \gamma^3 \mathrm{d} x = mc^2\cdot \gamma (u(x_2)) -mc^2\cdot \gamma(u(x_1))$

Darn, I'm on my phone, cannot see posted equations.

Best? Occam's razor kind of best.

 

[malawi_glenn]

Best? Occam's razor kind of best.

E = mc^2 is also not even the full story. It is a special case of something having zero speed as measured in a certain frame of reference.

 

[JLT]

E = mc^2 is also not even the full story. It is a special case of something having zero speed as measured in a certain frame of reference.

No aether, what was it- splitting light and bouncing back no matter how the thing is rotated with respect to the rotating earth that is flying around the sun.

I'm on vacation, just looking for book suggestions.

 

[malawi_glenn]

No aether, what was it- splitting light and bouncing back no matter how the thing is rotated with respect to the rotating earth that is flying around the sun.

I'm on vacation, just looking for book suggestions.

There are plenty of threads where people are asking for book recommendations regarding learning special relativity.
I can suggest the book by Morin https://scholar.harvard.edu/david-morin/special-relativity

 

[Dale]

TL;DR Summary: best derivation

Ok, joking aside, what is the best derivation of E= mc^2 ?

IMO, the best derivation is to derive the four-momentum and then define its norm to be mass.

 

[JLT]

Is there any relationship between:
E = m v^2
and E = Pressure*Volume?

PV = (density *Volume) * velocity^2
Pressure=density*velocity^2
almost starts looking like Bernoilli's equation.

 

[Dale]

Is there any relationship between:
E = m v^2
and E = Pressure*Volume?

The first one is wrong and the second one is right. Not sure if that is the sort of relationship you are referring to.

PV = (density *Volume) * velocity^2
Pressure=density*velocity^2
almost starts looking like Bernoilli's equation.

It looks like you are just throwing letters and equal signs randomly on a wall.

 

[JLT]

Sorry, on my phone without equation editor. Instead of E=m c^2, could be E=mv^2 where v is any constant velocity of any fluid. See original post, a chain rule, its a mass flow rate equation.

 

[robphy]

Is there any relationship between:
E = m v^2
and E = Pressure*Volume?

PV = (density *Volume) * velocity^2
Pressure=density*velocity^2
almost starts looking like Bernoilli's equation.

Of course, for dimensional reasons, one can see resemblances to possibly familiar equations.
While these may hint at connections from relativistic objects,
the appropriate path is to study the relativistic versions of those familiar equations:
e.g.
https://www.google.com/search?q=relativistic+kinetic+theory
https://www.google.com/search?q=relativistic+fluids
https://www.google.com/search?q=relativistic+statistical+mechanics

...but this is moving away from the original question.

 

[PeterDonis]

could be E=mv^2 where v is any constant velocity of any fluid. See original post, a chain rule, its a mass flow rate equation.

None of this has anything to do with $E = m c ^2$.

 

[Nugatory]

Occam's razor kind of best.

By that you mean the derivation that requires the fewest starting assumptions?
Most derivations start with two straightforwardly stated and universally applicable assumptions, namely Einstein’s postulates. From there it's just logic and a lot of high school math.

However it is a lot of that high school math. @malawi_glenn's post #2 in this thread follows the same path that Einstein and his contemporaries took to get to $E=mc^2$ from the postulates, but there's a fair amount of preliminary work required to get to where we can set up the integral. In #7 @Dale uses the norm of the four-vector, an approach that is much more aligned with modern thinking, but again there's a lot of preliminary work required to get to that four-vector formulation. I was tempted to post my "simplest" derivation: Set $p=0$ in the general relationship between mass, energy, and speed/momentum:$$E^2=(m_0c^2)^2+(pc)^2$$but of course that just invites the question, where did that relationship come from? And the answer is the same as for the other two: Start with Einstein's postulates and do a lot of high school math.

There's no way of getting to $E=mc^2$ without understanding all of special relativity, the same way that I can't fill the house with the delicious smell of a baking cake without actually baking a cake. That makes MalawiGlen's second answer, the one in #6, perhaps the most useful one.

 

[malawi_glenn]

the same way that I can't fill the house with the delicious smell of a baking cake without actually baking a cake

You can borrow my wife (or at least her perfume)

 

[Ibix]

You can borrow my wife (or at least her perfume)

...but can she derive $E=mc^2$?

 

[malawi_glenn]

...but can she derive $E=mc^2$?

She can memorize it for sure, but she can not understand a single step
But her perfume, I promise every day I get back from work I swear I think she has made newly baked cinnemon buns

 

[martinbn]

IMO, the best derivation is to derive the four-momentum and then define its norm to be mass.

What do you mean by derive the four-momentum?

 

[Dale]

What do you mean by derive the four-momentum?

I mean construct it and show that it is a four-vector.